The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

There are multiple solutions to this function

However, to validate the mean value theorem, the solution must fit within the interval . The only solution which does this is 

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of   on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

This is best found using a numerical solver

Though there are multiple options, the ones which fit within 

 are . All others, while solutions to the equation, do not satisfy the mean value theorem.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

There are multiple solutions of the form ; however, the only one which satisfies the mean value theorem and  falls within  is .

Rolle's Theorem states that if a real-valued function, , is continuous and differentiable on a closed interval , and if, then there must be some point, , in the open interval  that satisfies the equation .

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

Visually  is continous over 

Evaluate the function 

at its specified start and end points 

This satisfies  for Rolle's Theorem. Now differentiate:

The function is differentiable over the interval because its derivative is continuous; Rolle's Theorem is fully satisfied. 

Rolle's Theorem states that if a real-valued function, , is continuous and differentiable on a closed interval , and if, then there must be some point, , in the open interval  that satisfies the equation .

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

 is continuous from visual inspection. Evaluating the start and end points specified by 

Rolle's Theorem is not satisfied.

Rolle's Theorem states that if a real-valued function, , is continuous and differentiable on a closed interval , and if, then there must be some point, , in the open interval  that satisfies the equation .

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

The function  has no discontinuities, and looking at start/end points for the interval :

we see that .

However, the question becomes whether or not the function is differentiable over this interval.

The limit of  as  is different depending on if we approach zero from the left or the right. The derivative is not continuous, and the function is not differentiable over .

Rolle's Theorem is not satisfied.

Rolle's Theorem states that if a real-valued function, , is continuous and differentiable on a closed interval , and if, then there must be some point, , in the open interval  that satisfies the equation .

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

Looking at the function , there are no obvious discontinuities. Now consider its start and end points 

So far so good.

Now, the question is whether or not the function is differentiable.

Compare the derivative of these three functions at 

The function is differentiable as well. All of the criteria for Rolle's Theorem are satisfied. And, as it stands, there is just one point that satisfies , and we just evaluated it.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function  at the point 

The slope of the tangent is

The equation of a tangent line follows the form

, where m is the slope (just found), and b is a constant to ensure the line intercepts the original function.

So we currently have

Since this is where the tangent must intercept the function at :

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function  at .

The slope of the tangent is

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

Product rule: 

Note that u and v may represent large functions, and not just individual variables!

Taking the derivative of the function  at .

The slope of the tangent is